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How many half-lives will be required for 99.9% completion of 1st order reaction?

Option 1 -

3

Option 2 -

5

Option 3 -

8

Option 4 -

10

0 2 Views | Posted 4 weeks ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 weeks ago
    Correct Option - 4


    Detailed Solution:

    t 99.9 = 2.303 k log ? 100 100 - 99.9 = 3 × 2.303 k

    t 50 = 0.693 k

    t 99.9 t 50 = 3 × 2.303 × k k × 0.693 = 9 96 ≈ 10

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